k 1 ¼ Àb Æ
1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2
qx 2
2 Àr 2
r
:
ð3:137Þ
The solution a = 0 is unstable if Re(k 1 ) > 0, that is, if:
q [
2
x 2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r 2 þ ð2bÞ
2
q
:
ð3:138Þ
Since c does not appear in this condition, the stability of the zero solution does
not depend on the nonlinearity.
We then examine the stability of the two nonlinear solutions given by (3.129),
denoted here by a 1 and a 2 :
a
2
1 ¼
4
3c
ðr þ
ffiffiffiffiffi ffi
C 1
p Þ; a
2
2 ¼
4
3c
ðr À
ffiffiffiffiffi ffi
C 1
p Þ ;
ð3:139Þ
where
C 1
1
2
qx
2
2 À 2b
ð Þ
2 :
ð3:140Þ
The Jacobian (3.135) evaluated at these two singular points becomes:
Jð~ a; ~
wÞ ¼
0
1
2 ~ a
3
4 c~ a
2
À r
ð
Þ
À
3
2 c~ a
À2b
!
;
ð3:141Þ
where ~ a denotes either a 1 or a 2 , and where (3.128) has been used for eliminating
sinw and cosw. The eigenvalues of J(~ a, ~
w) are determined by the requirement that |J
(~ a, ~
w) – kI| = 0, which leads to a polynomial equation in k:
k
2
þ 2bk þ C 2 ¼ 0;
ð3:142Þ
with solutions
k ¼ Àb Æ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b
2
À C 2
q
;
ð3:143Þ
where
C 2
3
4
c~ a
2 3
4
c~ a
2
À r
:
ð3:144Þ
Unstable solutions are characterized by having at least one eigenvalue k with a
positive real part. By (3.143) a positive real part requires:
138
3 Nonlinear Vibrations: Classical Local Theory